Python 2, 62 bytes

lambda n:["%*s"%(i%2*2*n-n,"x"*min(i+1,n-i))for i in range(n)] 

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Uses x and space.

The rows are computed like this:

"%-5s" % "x" == "x " "%5s" % "xx" == " xx" "%-5s" % "xxx" == "xxx " "%5s" % "xx" == " xx" "%-5s" % "x" == "x " 

Using the %*s specifier to choose between n and -n.

APL (Dyalog Classic), 18 bytes

⎕a[↑⊢∘⌽\(⊂>⊢⌊⌽)⍳⎕] 

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outputs AB instead of *#

⎕ evaluated input n

⍳⎕ the vector 0 1 ... n-1

⊢⌊⌽ min (⌊) between themselves (⊢) and their reverse (⌽) - see trains

⊂>⊢⌊⌽ where is the ⍳ vector as a whole (⊂) less than each of its ⊢⌊⌽ - return a vector of boolean (0/1) vectors

⊢∘⌽\ reverse every other vector

↑ mix into a matrix

⎕a the uppercase English alphabet, 'AB...Z'

⎕a[ ] replace 0 1 with 'A' 'B'

Perl 6, 57 bytes

{map ^$^a: {fmt -$a*(-1)**$_=>\*x($_+1 min$a-$_): "%*s"}} 

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Uses the same method as Lynn's Python answer. Outputs using * and space.

MATL, 34 31 18 bytes

:t!>tPY|!"@X@oQ&P! 

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Uses 0 for * and 1 for #. Based on Lynn's Jelly answer.

Older answer, 31 bytes:

2/tk:wXk:Ph"X@ot~XHh@Gy-hHQ&PY" 

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Uses 1 for * and 0 for #.

 % implicit input, say 5 2/ % divide input number by 2 [2.5] tk % make a copy and floor that [2.5, 2] : % create range 1 to the floored value [2.5, [1, 2]] wXk % bring out the division result and this time ceil it % [[1, 2], 3] : % create range 1 to that [[1, 2], [1, 2, 3]] Ph % flip the last array and concatenate horizontally % [[1, 2, 3, 2, 1]] " % loop through the array X@o % Is the current loop index odd? 1 for odd, 0 for even t~ % duplicate and logical negate that XH % copy that value to clipboard H h % and concatenate the values ([1 0] on odd iterations, [0 1] on even) @ % push current value from array (say 2, then stack is [[0 1], 2) G % push input again y- % subtract current array value from input [[0 1], 2, 3] h % concatenate those two [[0 1], [2, 3]] H % get the stored value from clipboard H (1 for even iterations, 0 for odd) Q % increment that &P % flip the array in that dimension: in even iterations, this flips % across columns and hence inverts the two values. [[0 1], [3, 2]] % in odd iterations, it's a no-op Y" % run-length decoding - repeat the element from first array the number of times % specified in the second array % implicit loop end, implicit output 

Charcoal, 21 bytes

≔⮌…⁰ＮθＥθ⭆蛧⟦μλ⟧κ⌊⟦κι 

Try it online! Uses 0 and 1. Link is to verbose version of code and includes §*# which translates the output to the * and # in the question. Explanation:

 Ｎ Input number …⁰ Range from 0 ⮌ Reversed ≔ θ Assign to `q` Ｅθ Map over reversed range ⭆θ Map over reversed range and join §⟦μλ⟧κ Alternate between range and reversed range column ⌊⟦κι Minimum of range and reversed range row › Greater Implicitly print each row on its own line 

Jelly,  12  15 bytes

+3 fixing n=1 edge-case bug :(

R«Ṛ$‘r⁸ṬUÐe0YE? 

A full program accepting an integer which prints the output as defined in the OP using 0 and 1 for * and # respectively.

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How?

R«Ṛ$‘r⁸ṬUÐe0YE? - Main Link: integer, n R - range -> [1,2,3,4,...,n] $ - last two links as a monad: Ṛ - reverse -> [n,...,4,3,2,1] « - minimum (vectorises) -> [1,2,3,4,...,4,3,2,1] ‘ - increment (vectorises) -> [2,3,4,5,...,5,4,3,2] ⁸ - chain's left argument, n r - inclusive range (vectorises) -> [[2,3,...,n],[3,4,...n],[4,5,...n],[5,...n],...,[5,...n],[4,5,...n],[3,4,...n],[2,3,...,n]] Ṭ - untruth (vectorises) -> [[0,1,1,...,1],[0,0,1,1,...,1],[0,0,0,1,...,1],[0,0,0,0,1,...,1],...,[0,0,0,0,1,...,1],[0,0,0,1,...,1],[0,0,1,1,...,1],[0,1,1,...,1]] Ðe - apply to entries with even indices: U - upend -> [[0,1,1,...,1],[1,1,...,1],[0,0,0,1,...,1],[1,...,1,0,0,0,0],...] ? - if... E - ...condition: all equal? (only true when n=1, where we have [1,1]) 0 - ...then: zero Y - ...else: join with newline characters - implicit print 

Jelly, 15 bytes

r1«RR;ṬṖɗ€‘UÐeY 

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Full program.

* = 1
# = 0

Java 10, 145 bytes

n->{var r=new char[n][n];for(int j=0,k;j<n;++j)for(k=0;k<n;)r[j][k]=k++<(j<n/2?j%2<1?j+1:n+~j:j%2>0?j:n-j)?j%2<1?'*':'#':j%2>0?'*':'#';return r;} 

All the ternary makes it a bit messy, but it works fine. I tried flattening the nested loop, and various other things, but they only increased the byte count. Try it online here.

Ungolfed:

n -> { // lambda taking an integer as output and returning a char[][] var r = new char[n][n]; // the output array; we make use of Java 10's var here (replace with char[][] for another 4 bytes to make this work in Java 8) for(int j = 0, k; j < n; ++j) // iterate over the lines for(k = 0; k < n; ) // iterate over the j'th line r[j][k] = // set the current character k++ < // determine if we're in the first or second portion of the line: (j < n/2 ? // for the first half of the output: j%2 < 1 // on even lines ... ? j + 1 // ... print the first symbol j+1 times ... : n + ~j // ... on odd lines, print it n-j-1 times. : j%2 > 0 ? // for the second half of the output, on odd lines ... j : // ... print the first symbol j times ... n - j) // ... on even lines, print it n-j times. ? j%2 < 1 ? '*' : '#' // for the first part of the line, use '*' on even lines, '#' otherwise : j%2 > 0 ? '*' : '#'; // for the second part of the line, use '*' on odd lines, '#' otherwise return r; // return the completed array } 

Java 8 11, 179 127 bytes

n->{String r="",a,b;for(int j=0;j<n;b="#".repeat(j<n/2?n+~j:j),r+=(j++%2<1?a+b:b+a)+"\n")a="*".repeat(j<n/2?j+1:n-j);return r;} 

Try it online here (TIO does not have Java 11 yet, so this uses a custom method that results in the same byte count as String#repeat()).

Thanks to Kevin Cruijssen for golfing a whopping 52 bytes!

Ungolfed:

n -> { // lambda taking an int argument and returning a String String r = "", // the output String a, // temporary String containing the '*'s b; // temporary String containing the '#'s for(int j = 0; j < n; // loop over the lines b = "#".repeat( // repeat the '#' character ... j < n/2 ? n + ~j // ... n-j-1 times in the first half of the output ... : j), // ... j times in the second half r += (j++ % 2 < 1 ? a + b : b + a) + "\n") // assemble the j'th line and append it to the output: on even lines, the '*'s go first; on odd lines, the '#'s go first a = "*".repeat( // repeat the '*' character ... j < n/2 ? j + 1 // ... j+1 times in the first half of the output ... : n - j); // n-j times in the second half return r; // return the completed output } 

Lua,  148  133 Bytes

function(n)t,a,b={},".","#"for i=1,n do r=i<n/2+1 and i or-~n-i s=a:rep(r)..b:rep(n-r)t[i]=i%2<1 and s:reverse()or s end return t end 

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-15 bytes thanks to @KevinCruijssen and @JoKing.

function(n) t = {}; a = "."; b = "#" -- initialize variables, output is in table -- strings are needed in variables for -- the str:rep and str:reverse syntax for i = 1, n do -- build the rows of the table r = i<=(n+1)/2 and i or n-i+1 -- logic used to count up then down str = a:rep(r)..b:rep(n-r) -- append correct number of '.'s, fill -- in the rest with '#'s t[i]=i%2==0 and str:reverse() or str -- logic used to control reversing end return t -- return table end 

Kotlin, 86 bytes

{n:Int->for(i in 0..n)println("%${i%2*2*n-n}s".format("x".repeat(Math.min(i+1,n-i))))} 

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Uses x and space as output symbols. Derivates from Lynn Answer

Perl 5 + -pa -MPOSIX -M5.010, 58 bytes

say+sort{$|--}1x abs,2x("@F"-abs)for(1..ceil$_/=2),-$_..-1 

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C (gcc), 104 99 bytes

h,j,k;f(i){char s[h=i++];for(j=0;k=++j>i/2?i-j:j,j<i;memset(memset(s,42,h)+j%2*k,35,h-k),puts(s));} 

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C (gcc), 118 108 bytes

This one isn't going to win, but it's a different approach (or at least, I think so!) Instead of doing string manipulations, I make use of the fact that \$10^x-1\$ over \$[1..n] = \{9, 99, 999, ...\}\$, which then can be multiplied to get the appropriate pattern; printf() then does the zero-padding for right-justification.

Sadly, int only has sufficient range to do up to 9 digits (on 32-bit platforms), so you need to go to long for larger patterns; a language that natively does MP arithmetic might be able to use this for something.

Thanks to ceilingcat for the suggestion.

h,j,k;p(h){h=h?10*p(--h):1;}f(i){for(j=0,h=i++;k=++j>i/2?i-j:j,j<i;printf("%0*d\n",h,~-p(k)*p(j%2*(h-k))));} 

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Proof of concept that this works with MP arithmetic:

C# (Mono C# compiler), 187 165 bytes

(143 bytes + 22 bytes for the using System.Numerics; in the header)

q=>{var r="";for(int j=0,h=q+1,k;j<q;r+=((BigInteger.Pow(10,k)-1)*BigInteger.Pow(10,j%2*(q-k))).ToString("D"+q)+"\n")k=++j>h/2?h-j:j;return r;} 

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Vim, 99 keystrokes

It's always interesting to try to do vim with input arguments. It's very unnatural, so it's not going to be amazingly short. There probably are other good approaches to this.

Input is assumed to be by itself in a buffer. Registers are assumed to be empty. Editor is assumed to be tall enough to contain the result without scrolling (this could technically be avoided at the cost of some keystrokes).

"nD@ni<cr><esc>MmaGddM <c-v>'aI*<esc>qwgvjokoI*<esc>@wq@w<esc> :set ve=all<cr>@nlh<c-v>@nkr# :%s/ /#/g<cr>o<esc> 2Gqqdt#$p2j0@qq@q 

Explanation

 | Buffer state (odd and even case): | 5 6 "nD read input into register n @ni<cr><esc> add n newlines MmaGddM<c-v>'a visual block select center row(s) I*<esc> prepend a column of * qw record macro w gvjoko expand selection up and down I*<esc> @w recurse q @w<esc> run macro w and exit visual block select | Buffer state: | * * | ** ** | *** *** | ** *** | * ** | * :set ve=all<cr> move anywhere! @nlh<c-v>@nkr# add last column of #s | Buffer state: | * # * # | ** # ** # | *** # *** # | ** # *** # | * # ** # | * # :%s/ /#/g<cr> replace spaces with # | Buffer state: | *#### *##### | **### **#### | ***## ***### | **### ***### | *#### **#### | *##### o<esc>2G prep and jump to line 2 qqdt#$p2j0@qq@q (effectively) flip every other onward | Buffer state: | *#### *##### | ###** ####** | ***## ***### | ###** ###*** | *#### **#### | #####* 

And in base64, with actual characters (put input in input and keystrokes in keys and run using vim -u NONE -s keys input)

Im5EQG5pDRtNbWFHZGRNFidhSSobcXdndmpva29JKhtAd3FAdxs6c2V0IHZlPWFsbA1AbmxoFkBua3IjOiVzLyAvIy9nDW8bMkdxcWR0IyRwMmowQHFxQHE= 

R, 75 bytes

function(n)outer(1:n,1:n,function(x,y,a=x<y|x>n-y+1)+ifelse(x%%2,a,rev(a))) 

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• Inspired by @Lynn answer
• function getting n as parameter and returning a matrix of 0/1 where 0 corresponds to '*' and 1 corresponds to '#'

K (ngn/k), 22 bytes

{"*#"i|:/'i>/:i&|i:!x} 

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{ } function with argument x

!x the list (0;1;...;x-1)

i: assign to i

i&|i minima (&) of i and its reverse (|)

i>/: compare with greater than (>) i against each element from the list on the right (/:) - return a boolean matrix (list of lists)

i|:/' for each (') j in i, reverse (|: - we need the : to force | to be unary) the corresponding element j times (n f/ x applies f n times on x). Effectively, reverse every other row.

"*#" use matrix elements as indices in the string "*#"

Python 3, 251 bytes

n = int(input()) for i in range(1,n+1): if i%2 != 0: for j in range(1,n-i+2): print("*",end=" ")   for j in range(1,i+1): print("#",end=" ")   if i%2 == 0: for j in range(1,n-i+2): print("*",end=" ")   print() 

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